• LA 3263 好看的一笔画 欧拉几何+计算几何模板


    题意:训练指南260

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    #include <cmath>
    using namespace std;
    
    struct Point {
        double x, y;
        Point(double x = 0, double y = 0) : x(x) , y(y) { }
    };
    
    typedef Point Vector;
    
    Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }
    Vector operator - (Vector A, Vector B) { return Vector(A.x-B.x, A.y-B.y); }
    Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
    Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
    
    bool operator < (const Point& a, const Point& b) {
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    }
    
    const double eps = 1e-10;
    int dcmp(double x) {
        if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
    }
    
    bool operator == (const Point& a, const Point& b) {
        return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
    }
    
    double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
    double Length(Vector A) { return sqrt(Dot(A, A)); }
    double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
    
    double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
    double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); }
    
    Vector Rotate(Vector A, double rad) {
        return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad) );
    }
    
    Vector Normal(Vector A) {
        double L = Length(A);
        return Vector(-A.y/L, A.x/L);
    }
    
    Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
        Vector u = P - Q;
        double t = Cross(w, u) / Cross(v, w);
        return P + v * t;
    }
    
    double DistanceToLine(Point P, Point A, Point B) {
        Vector v1 = B-A, v2 = P - A;
        return fabs(Cross(v1,v2) / Length(v1));
    }
    
    double DistanceToSegment(Point P, Point A, Point B) {
        if(A==B) return Length(P-A);
        Vector v1 = B - A, v2 = P - A, v3 = P - B;
        if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
        else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
        else return fabs(Cross(v1, v2)) / Length(v1);
    }
    
    Point GetLineProjection(Point P, Point A, Point B) {
        Vector v = B - A;
        return A + v * ( Dot(v, P-A) / Dot(v, v) );
    }
    
    bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) {
        double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
                c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
        return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
    }
    
    bool OnSegment(Point p, Point a1, Point a2) {
        return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
    }
    
    double ConvexPolygonArea(Point* p, int n) {
        double area = 0;
        for(int i = 1; i < n-1; i++)
            area += Cross(p[i] - p[0], p[i + 1] - p[0]);
        return area / 2;
    }
    
    const int maxn = 300 + 10;
    Point P[maxn], V[maxn*maxn];
    
    Point p[305],v[305*305];
    int main()
    {
        int n,cas=0;
        while(~scanf("%d",&n)&&n)
        {
            cas++;
            for(int i=1;i<=n;i++)
               {
                    scanf("%lf %lf",&p[i].x,&p[i].y);
                    v[i]=p[i];
               }
            n--;
    
            int cnt=n+1;
            for(int i=1;i<=n;i++)
              for(int j=i+1;j<=n;j++)
               if(SegmentProperIntersection(p[i],p[i+1],p[j],p[j+1]))
                    v[++cnt]=GetLineIntersection(p[i],p[i+1]-p[i],p[j],p[j+1]-p[j]);
    
    
            sort(v+1,v+cnt+1);
            int vnum=unique(v+1,v+cnt+1)-(v+1);
            //for(int i=1;i<=vnum;i++)
              //  printf("v %d:%f %f
    ",i,v[i].x,v[i].y);
            int e=n;
            //cout<<"ori   "<<e<<endl;
            for(int i=1;i<=vnum;i++)
                for(int j=1;j<=n;j++)
                  if(OnSegment(v[i],p[j],p[j+1]))
                      e++;
           //cout<<"vnum:"<<vnum<<"  e:"<<e<<endl;
            printf("Case %d: There are %d pieces.
    ",cas,e+2-vnum);
        }
        return 0;//v+f-e=2;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/smilesundream/p/5267183.html
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